Problem: $(2\sin y + 1)\,\dfrac{dy}{dx}=4$ and $y(0)=\pi/2$. What is $x$ when $y=\pi$ ? $x=~$
The differential equation is separable. What does it look like after we separate the variables? $(2\sin y + 1)\,dy= 4\,dx$ Let's integrate both sides of the equation. $\int (2\sin y + 1)\,dy= \int 4\,dx$ What do we get? $-2\cos y + y =4x+C$ What value of $C$ satisfies the initial condition $y(0)=\pi/2$ ? Let's substitute $x=0$ and $y=\pi/2$ into the equation and solve for $C$. $\begin{aligned} -2\cos \dfrac\pi2 + \dfrac\pi2 &=4\cdot0+C\\ \\ \\ -2\cdot0 + \dfrac\pi2 &=C\\ \\ \\ C&=\dfrac\pi2 \end{aligned}$ Now use this value of $C$ to find $x$ when $y=\pi$. $\begin{aligned} -2\cos \pi + \pi&=4x+\dfrac\pi2\\ \\ \\ 2 + \pi&=4x+\dfrac\pi2\\ \\ \\ 4x&=2+\dfrac\pi2\\ \\ \\ x&=\dfrac12+\dfrac\pi8 \end{aligned}$